This puzzle replaces all numbers with other symbols.

Your job, as the title suggests, is to find what value fits in the place of $\bigstar$. To get the basic idea, I recommend you solve Puzzle 1 first.

All symbols follow these rules:

  1. Each numerical symbol represents integers and only integers. This means fractions and irrational numbers like $\sqrt2$ are not allowed. However, negative numbers and zero are allowed.
  2. Each symbol represents a unique number. This means that for any two symbols $\alpha$ and $\beta$ in the puzzle, $\alpha\neq\beta$.
  3. The following equations are satisfied (this is the heart of the puzzle): $$ \text{I. }a^a=a \\ \space \\ \text{II. }a+a+a=b \\ \space \\ \text{III. }c<d<c^b\times(b-c) \\ \space \\ \text{IV. }a\times(c+d)=e\times e \\ \space \\ \text{V. }f^g=g^f \\ \space \\ \text{VI. }c\times c+d\times d<(f+g)\times e\times e \\ \space \\ \text{VII. }c-b<h<(e+a)^f \\ \space \\ \text{IIX. }h<\bigstar <e-h $$

What is a Solution?

A solution is a value for $\bigstar$, such that, for the group of symbols in the puzzle $S_1$ there exists a one-to-one function $f:S_1\to\Bbb Z$ which, after replacing all provided symbols using these functions, satisfies all given equations.

Can you prove that there is only one possible value for $\bigstar$, and find that value?

Good luck!

Side Note: to get $\bigstar$ use $\bigstar$, and to get $\text^$ use $\text^$

Previous puzzles:

Introduction: #1 #2 #3 #4 #5 #6 #7

Inequalities: #8

Next Puzzle


1 Answer 1


We must have



V is true when f=g, of course, but that won't do; also when {f,g}={2,4} or when {f,g} = {-2,-4}; and those are all the integer solutions to V. So f,g are 2,4 in some order or -2,-4 in some order. The latter is impossible because VI says positive < (f+g) . positive, which is impossible if f+g is negative.


I is true only when a is -1 or +1. (You might argue for 0, but II forbids that since then we'd have b=a.) By II, b is then -3 or +3.

Let's look first at the case where

a=1, b=3. Then III says c < d < c^3(3-c). Can we have c>0? No! The LHS is positive so the RHS had better be too, so c<3. But 1,2,3 are already taken, contradiction. Can we have c=0? No! We get 0 < d < 0. What if c<0? Then the bounds on d are the wrong way around, because c^3(3-c) is always more negative than c. (Because it's c times c^2(3-c); the latter is the product of two positive factors, at least one of which must be >1.)


a=-1, b=-3. Now III says c < d < c^-3.(-3-c). If c>0 then that RHS is negative, so we have positive < something < negative, contradiction; if c=0 then it says 0 < d < 0, contradiction, so c<0. Then either c=-2 or c<-3. In the former case we have -2 < d < -1/8.(-3 - -2) = 1/8 which (since -1 is already taken) requires d=0. Then IV says -1 . (-2+0) = e^2 which is impossible. So c<-3. Then c < d < 1/c^3 . (-3-c) and the RHS always lies between -1 and 0. So d is negative (and must in fact be <= -2, since -1 is already taken) but less negative than c. IV then says that -(c+d) = e^2.

Note now that

f+g=6 so VI says c^2+d^2 < 6e^2 = -6(c+d). That's quite constraining. In particular it implies -(c+d) < 12. (Because c^2+d^2 >= (c+d)^2/2, so (c+d)^2/2 < -6(c+d); dividing by the positive quantity -(c+d)/2 we get -(c+d) < 12.) This thing has to be a square. Which square? Well, c <= -4 and d <= -2, so 6 = 2+4 <= -(c+d) <= 12. The only square in this range is 9.


the only ways to make 9 with our given constraints so far are (-7,-2) and (-5,-4). We can readily check that the former violates VI. So c=-5 and d=-4. And we have e^2=9 so e must be 3. (-3 is already taken.)

Our remaining conditions now look like

-2 < h < 2^f and h < $\bigstar$ < 3-h. The latter implies 3-h >= h+2 or h <= 1/2, so in fact we must have h=0 and then $\bigstar$ is either 1 or 2 -- but 2 is already taken, so $\bigstar$ is 1.

We have nailed down

specific values for all variables except f,g, which could be either 2,4 or 4,2.

  • $\begingroup$ You got the right value, which is nice, but some things are missing: 1. "those are all the solutions to Eq.V". Are you sure about that? 2. on your first case with Eq.III why must LHS be positive? 3. the second row of the "Note now that" block is confusing me. How did you get those conclusions? $\endgroup$
    – NODO55
    Commented Feb 25, 2018 at 15:48
  • $\begingroup$ 1 and 2 are genuine oversights; I'm in the process of fixing them. 3 could use another few words of explanation, which I'm adding at the same time. [EDITED to add:] Now done. $\endgroup$
    – Gareth McCaughan
    Commented Feb 25, 2018 at 18:21
  • $\begingroup$ this is a great improvement, however I still don't get how you came to the conclusion of the now third row of the "Note now that" block (it was on the second row but oh well). Good job though! $\endgroup$
    – NODO55
    Commented Feb 25, 2018 at 18:39
  • $\begingroup$ I've made it more explicit; how do you find it now? (Perhaps I've misidentified the bit that wasn't clear, in which case please accept my apologies.) $\endgroup$
    – Gareth McCaughan
    Commented Feb 25, 2018 at 19:36
  • $\begingroup$ The bit you edited was very clear. The bit I don't understand is how you concluded that -(c+d)>=-(-4-2) $\endgroup$
    – NODO55
    Commented Feb 25, 2018 at 19:41

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